Ejaifeobuks wrote: Let $a,b,c$ be real numbers satisfying $ab+bc+ca=1.$ Show that $$\frac{|a-b|}{1+c^2} +\frac{|b-c|}{1+a^2} \geq\frac{|c-a|}{1+b^2}$$ Write the denominators as$$1+a^2=a^2+ab+bc+ca=(a+b)(a+c)$$Now, multiply both sides by $|(a+b)(b+c)(c+a)|$, and use triangle inequality:$$|a^2-b^2|+|b^2-c^2| \geq |c^2-a^2|$$Nimo TST 2019

Ejaifeobuks wrote: Let $abc$ be real numbers satisfying $ab+bc+ca=1$. Show that $\frac{|a-b|}{|1+c^2|}$ + $\frac{|b-c|}{|1+a^2|}$ $>=$ $\frac{|c-a|}{|1+b^2|}$ $$\frac{|a-b|}{|1+c^2|}+\frac{|b-c|}{|1+a^2|}=\frac{|(a+b)(a-b)|+|(b+c)(b-c)|}{|a+b||b+c||c+a|}$$$$\geq\frac{|(a+b)(a-b)+(b+c)(b-c)|}{|a+b||b+c||c+a|}=\frac{|c-a|}{|a+b||b+c|}=\frac{|c-a|}{|1+b^2|}.$$